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## Comment on

Triangles - Part II## At 7:21, if we draw an

## Good question.

Good question.

The answer is no. In order for the diagonal to bisect the top angle, the non-equal side must be the base.

Try it with a 45-45-90 triangle, and you'll see what I mean.

## Hi Brent,

A little confused with this. Suppose if the isosceles triangle is ABC with sides AB=AC, then will the altitude drawn from angle A bisect angle BAC in two equal parts? I thought it is only a bisector segment dividing side BC in two equal parts.

Thanks

## Yes, the altitude drawn from

Yes, the altitude drawn from point A WILL bisect angle BAC in two equal parts.

Think of it this way (using the diagram you describe with point D representing the point where the altitude intersects side BC)

We get two triangles: ∆ADB and ∆ADC

First, recognize that ∠ABD = ∠ACD, since those are the equal angles in ∆ABC

Second, recognize that ∠ADB = ∠ADC = 90°, since AD is the ALTITUDE

Third, since triangles ∆ADB and ∆ADC already have two angles in common, the third angles must also be equal. That is ∠BAD = ∠CAD

Does that help?

Cheers,

Brent

## Amazing. Without an inkling

Is a bisector drawn always perpendicular to the line it bisects or this is the property applicable only in case of Isosceles and Equilateral triangle?

Thanks.

## The bisector is perpendicular

The bisector is perpendicular ONLY IF the triangle is either isosceles and equilateral.

## Sir is there any easy

https://gmatclub.com/forum/in-the-figure-above-point-o-is-the-center-of-the-circle-and-107874.html

## Have you seen Karishma's

Have you seen Karishma's solution: https://gmatclub.com/forum/in-the-figure-above-point-o-is-the-center-of-...

If you have any questions about her solution, let me know and we can go from there.

## sir, i am having doubt in

https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-side-ac-and-82671.html

## I have provided a detailed

I have provided a detailed response here: https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-si...

## One of the statements from

"(1) x^2 + y^2≠ z^2"

Does this statement translate into the triangle being an equilateral triangle?

## Question link: https:/

Question link: https://gmatclub.com/forum/is-the-area-of-the-triangular-region-above-le...

IF it were the case that x² + y² = z², then we'd know that the triangle is an RIGHT TRIANGLE with z as the hypotenuse.

Since we're told that x² + y² ≠ z², then we know that z is not the hypotenuse of a right triangle.

ASIDE #1: This does not necessarily mean that the given triangle is not a right triangle. It could be the case that side x is the hypotenuse of the triangle, in which case, we'd get y² + z² = x²

ASIDE #2: It COULD also be the case that the given triangle is an equilateral triangle, but it could also not be an equilateral triangle.

Does that help?

Cheers,

Brent

## Hi Brent

Helps perfectly thank you, I saw in a later video this topic is discussed in more depth.

Regards

## Hi Brent,

Could you please present your answer on this?

https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-side-ac-and-82671.html

The question introduces properties of Midsegements, something i haven't come across in our course module.

Thanks.

## You bet!

You bet!

Here's my step-by-step solution: https://gmatclub.com/forum/in-triangle-abc-point-x-is-the-midpoint-of-si...

Cheers,

Brent

## Question Link: https:/

Would you mind explaining the solution. Specifically I would like to know the approach on how the deduce or crack the given info. I seem to understand but it when encountered again, I can't get it right

## Hi emailme,

Hi emailme,

Here's my step-by-step solution: https://gmatclub.com/forum/in-the-figure-above-point-o-is-the-center-of-...

Cheers,

Brent

## Add a comment